Hi,
I see where I am going wrong. Maybe I need give more thought before
coming to the list for help.
Sorry.
Cheers,
Omprakash.V
omprakash(_dot_)v(_at_)pol
aris.co.in To:
<xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
cc: (bcc: omprakash.v/Polaris)
04/01/2005 Subject: [xsl] namespaces in
xslt
05:24 PM
Please respond
to xsl-list
Hi,
Iam having diffculty following what's happening:
I have xml like this
<table>
<th>
<root>
<tr>
<slideshow xmlns="urn:hl7-org/v3" title="Sample Slide Show" date="Date of
publication" author="Yours Truly">
<td><a href="file:///N65538"/>
</td>
</slideshow>
</tr>
</root>
</th>
</table>
And my xsl:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl
="http://www.w3.org/1999/XSL/Transform" xmlns="urn:hl7-org/v3" >
<xsl:output method="xml" indent="yes" />
<xsl:template match="/">
<xsl:for-each select="//root/*[1]">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:apply-templates select="@*|node()">
<xsl:with-param name="rootnode" select="$rootnode"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
When I select //own:root/*[1] in the <xsl:for-each>, I get the following
<?xml version="1.0" encoding="UTF-8"?>
<tr xmlns="http://XYZ"/>
but when I make it //root/*[2], I get nothing. But if you see, the
slideshow is really at position 2 so I need to use 2.
Cheers,
Omprakash.V
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